/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Integrate \(g(x, y, z)=x \sqrt{y... [FREE SOLUTION] | 91Ó°ÊÓ

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Integrate \(g(x, y, z)=x \sqrt{y^{2}+4}\) over the surface cut from the parabolic cylinder \(y^{2}+4 z=16\) by the planes \(x=0, x=1\) and \(z=0\)

Short Answer

Expert verified
Evaluate the triple integral by integrating successively over z, y, and x.

Step by step solution

01

Understanding the Surface

The given surface is a part of the parabolic cylinder described by the equation \( y^2 + 4z = 16 \). This surface is bounded by the planes \( x = 0 \), \( x = 1 \), and \( z = 0 \). To clarify, this means we are looking at a "sheet" cut out from the cylinder in the specified domain. The surface extends along the x-axis from 0 to 1.
02

Finding the Limits of Integration

From the equation \( y^2 + 4z = 16 \), we can express \( z \) in terms of \( y \) as \( z = \frac{16 - y^2}{4} \). The intersection of the cylinder with the plane \( z=0 \) occurs at \( y^2 = 16 \), giving the bounds \( y = -4 \) to \( y = 4 \). Thus, the integration limits are \( x = 0 \) to \( x = 1 \), \( y = -4 \) to \( y = 4 \), and \( z = 0 \) to \( z = \frac{16 - y^2}{4} \).
03

Setting Up the Triple Integral

We set up the triple integral to evaluate the integral over the surface. The integral is \( \int_{0}^{1} \int_{-4}^{4} \int_{0}^{\frac{16-y^2}{4}} g(x,y,z) \ dz \, dy \, dx \). Substitute the expression for \( g(x, y, z) = x \sqrt{y^2 + 4} \) into the integral.
04

Integrating with respect to z

Integrate the expression \( x \sqrt{y^2 + 4} \) with respect to \( z \) from 0 to \( \frac{16 - y^2}{4} \). The function does not depend on \( z \), so integration simplifies to multiplying the integrand by \( \frac{16 - y^2}{4} \).
05

Integrating with respect to y

After integrating with respect to \( z \), we have: \( \int_{-4}^{4} x \sqrt{y^2 + 4} \frac{16 - y^2}{4} \, dy \). Evaluate this integral for each fixed \( x \). Note that this can involve substitution or finding standard integral forms to solve.
06

Integrating with respect to x

Finally, integrate the result from the previous step with respect to \( x \) from 0 to 1 :: \( \int_{0}^{1} \). Since the previous step results in a constant coefficient times x, this integration is straightforward.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabolic Cylinder
A parabolic cylinder is a three-dimensional shape created by extending a parabola along a line, often the x-axis, to form a surface. Unlike a standard cylinder, which is based on a circle, a parabolic cylinder takes its cross-sectional shape from a parabola. In our exercise, the parabolic cylinder is described by the equation \( y^2 + 4z = 16 \). This specific equation gives us all the points \((y, z)\) that lie on the surface for any fixed value of \(x\).
The equation can be rearranged to solve for \(z\), showing the relationship between \(y\) and \(z\):
  • When \(y\) changes, \(z\) adjusts according to \(z = \frac{16-y^2}{4}\).
  • The surface extends infinitely in one direction but is bounded by given limits in this exercise.
In practical terms, the parabolic cylinder in our task is truncated by planes, making it easier to visualize and integrate over. You can imagine it as a curved wall within specified boundaries.
Triple Integration
Triple integration involves calculating the integral of a function in three dimensions, usually to determine the volume under a surface or the total of some property, like mass or charge, contained within a 3D space. In the context of this problem, we integrate a function \(g(x, y, z) = x \sqrt{y^2 + 4}\) over a defined region bounded by a parabolic cylinder and planes.
The process involves several steps:
  • First, integrating with respect to one variable, often starting with \(z\), because it is the innermost integration.
  • Then integrating with respect to another variable, \(y\), while keeping \(x\) constant.
  • Finally, integrating with respect to \(x\).
Each integration step narrows down the dimensions of our problem until we finally reach a single numerical answer, representing the total within that region. Triple integration requires careful setup of limits, ensuring every dimension is accounted for correctly.
Limits of Integration
When setting up an integral, limits of integration are essential for defining the region over which you integrate. These limits tell you where integration starts and stops for each variable. In some cases, these are constants, but in our problem, certain limits are functions of other variables.
For the given surface bounded by a parabolic cylinder, the limits were:
  • For \(x\), the limits are simple constants from 0 to 1, representing the portion of the x-axis.
  • For \(y\), the limits range from -4 to 4, which are found by examining where the parabolic cross-section intersects with the plane \(z=0\), when \(y^2 = 16\).
  • For \(z\), the upper limit changes with \(y\): \(z = \frac{16 - y^2}{4}\). This shows that as \(y\) varies, the height of the cylinder at that point changes.
Understanding these limits is crucial for accurately setting up and computing the triple integral. They define the shape and size of the region you are integrating over, ensuring you capture the entire volume enclosed by the boundaries.

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Most popular questions from this chapter

Let \(\mathbf{F}\) be a differentiable vector field defined on a region containing a smooth closed oriented surface \(S\) and its interior. Let \(\mathbf{n}\) be the unit normal vector field on \(S .\) Suppose that \(S\) is the union of two surfaces \(S_{1}\) and \(S_{2}\) joined along a smooth simple closed curve \(C .\) Can anything be said about $$ \iint_{S} \nabla \times \mathbf{F} \cdot \mathbf{n} d \sigma ? $$ Give reasons for your answer.

Let \(\mathbf{n}\) be the outer unit normal (normal away from the origin) of the parabolic shell $$ \text { S: } 4 x^{2}+y+z^{2}=4, \quad y \geq 0 $$ and let $$ \mathbf{F}=\left(-z+\frac{1}{2+x}\right) \mathbf{i}+\left(\tan ^{-1} y\right) \mathbf{j}+\left(x+\frac{1}{4+z}\right) \mathbf{k} $$ Find the value of $$ \iint_{S} \nabla \times \mathbf{F} \cdot \mathbf{n} d \sigma $$

Hyperboloid of one sheet a. Find a parametrization for the hyperboloid of one sheet \(x^{2}+y^{2}-z^{2}=1\) in terms of the angle \(\theta\) associated with the circle \(x^{2}+y^{2}=r^{2}\) and the hyperbolic parameter \(u\) associated with the hyperbolic function \(r^{2}-z^{2}=1 .\) (See Section \(7.8,\) Exercise \(84 . . )\) b. Generalize the result in part (a) to the hyperboloid \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)-\left(z^{2} / c^{2}\right)=1\)

Flux of a constant field Show that the outward flux of a constant vector field \(\mathbf{F}=\mathbf{C}\) across any closed surface to which the Divergence Theorem applies is zero.

Zero circulation Let \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2} .\) Show that the clockwise circulation of the field \(\mathbf{F}=\nabla f\) around the circle \(x^{2}+y^{2}=a^{2}\) in the \(x y\) -plane is zero a. by taking \(\mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi,\) and integrating \(\mathbf{F} \cdot d \mathbf{r}\) over the circle. b. by applying Stokes' Theorem.

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